Question

For a steady flow, the velocity field is $\overrightarrow{V}=(-x^2+3y)\hat{i}+\left(2xy\right)\hat{j}$. The magnitude of the acceleration of a particle at (1, -1) is

• A. 2
• B. 1
• C. $2\sqrt{5}$
• D. 0

Answer 1

The correct option is C $2\sqrt{5}$

Given velocity field,
$\overrightarrow{V}=(-x^2+3y)\hat{i}+\left(2xy\right)\hat{j}$
where $u=-x^2+3yandv=2xy$
The acceleration components along x and y-axis.
$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$
$and{a}_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$
$a_x=(-x^2+3y)\times (-2x)+2xy\times 3$
$=2x^3-6xy+6xy=2x^3$
$and{a}_y=(-x^2+3y)\times 2y+2xy\times 2x$
$=-2y{x}^2+6y^2+4x^{2}y=2y{x}^2+6y^2$
At point(1,-1)
$a_x=2$
$and{a}_y=2\times (-1)\times 1+6\times (-1{)}^2$
=-2+6=4
$\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}=2i+4j$
Resultant acceleration,
$a=\sqrt{4+16}=\sqrt{20}$
$=\sqrt{4\times 5}=2\sqrt{5}$